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**Example 2- Potential of an electric dipole**

Let’s calculate the potential of an electric dipole. Potential of an electric dipole.

As you recall, dipole is a point charge system which consists of two point charges with equal magnitudes and opposite signs, separated from one another by a very small distance. Let’s say we’re interested with the potential of such a system at a certain distance, *r*, away from the center of a dipole. Let’s say the distance between the positive charge and the point of interest is *r*+, and the distance between the negative charge and the point of interest is denoted with *r*-.

As you see, I’m not drawing any vectors, no directions, because I’m dealing with potential and potential is a scaler quantity. It’s not a vectoral quantity. It doesn’t have any directional properties. It is just electric potential energy per unit charge, and those two quantities are all scalers.

Well, to be able to calculate the potential of this system, we will first calculate the potential of each one of these charges at the point of interest, so *V*1 is going to be equal to the potential of a point charge is the charge divided by 4*π* *ε*0 times its distance to the point of interest, and that is *r*+ for this charge. Let’s call this one as *V*+. And for the negative charge, *V*-, we will have –*q* over 4*π* *ε*0 *r*-.

Total potential is the direct sum of these two scaler quantities, *V*+ plus *V*-, and therefore, *V* will be equal to, if we add these two quantities we will have *q* over 4*π* *ε*0 *r*+, and the other one is minus, so –*q* over 4*π* *ε*0 *r*-. We can write it in more compact form by taking common quantities parenthesis, which is in this case, *q*, in the numerator, and 4*π* *ε*0 in the denominator. This will be multiplied by 1 over *r*+ and minus 1 over *r*-.

Let’s go ahead and have common denominator in the parenthesis. Therefore, the potential will be equal to *q* over 4*π* *ε*0, *r*– minus *r*+ divided by *r*+ times *r*-. And this will be, basically, the answer. Knowing the distances of *r*– and *r*+ and the magnitude of the dipole, we can easily calculate the potential that it generates at that location.

Here now we’re going to look at an approximate case. Let’s denote this angle as *θ*, which is the angle that *r* is making with the dipole, and we will look at a special case and that is the case of that the point of interest’s distance to the dipole, which is *r*, is much, much greater than the dipole separation distance *d*. If this is the case, then we can visualize these three lines almost to be parallel to one another, and this angle being approximately equal to *θ*. If we just draw a line, or take the projection of this point along *r*-, then this distance is going to be equal to *r*– minus *r*+.

Furthermore, if this angle is approximately equal to *θ*, as you can see, we end up with formation of a right triangle over here, with this angle being 90 degrees, and the hypotenuse of this triangle is the dipole separation distance, *d*. In that case, *r*– minus *r*+, which is this distance, becomes approximately equal to *d* times cosine of *θ*.

Furthermore, for such a point which is very, very far away from the dipole in comparing to the separation distance, we can also approximately assume that these three distances are about the same. So we can write these expressions for *r* much, much greater than *d* distance as *r*– minus *r*+ is equal to, approximately equal to *d* cosine of *θ*, and *r*-, which is approximately equal to *r*+ and that is approximately equal to *r*. In that case, *r*– r plus approximately becomes equal to *r*2.

This enables us to write down this approximate expression for this specific case, and the potential becomes *q* over 4*π* *ε*0. So instead of *r*– minus *r*+, we will have d cosine of *θ*, and in the denominator, *r*– times *r*+ will give us approximately *r*2.

As you recall, when we were calculating the electric field of a dipole, we said that the charge of the dipole, the magnitude of the charge of the dipole, and the separation distance, *d*, these are unique properties of the specific dipole, and we had a special name for this product. We called it “magnitude of the electric dipole moment vector”, and we denoted this with *P*. In terms of electric dipole moment vector magnitude, then the potential can be expressed as *P* times cosine of *θ* over 4*π* *ε*0 *r*2.

When we look at this expression, we see that it takes its maximum and minimum value, depending upon this angle, *θ*, which is basically the orientation of this point, *P*, relative to the dipole, and we can easily see that for *θ* is equal to 90 degrees, cosine of 90 is 0, and this corresponds to the equatorial plane. And the potential of the dipole becomes equal to 0 for this case. That can be interpreted as charge lying on this plane always equidistant from the positive and the negative charges which make up the dipole so that the scaler potentials set up by each charge cancel each other.

Therefore, if you consider the equatorial plane, which is the plane passing through the center of the dipole, this plane, and perpendicular to the dipole, every point on this plane will be equidistant away from the positive charge and the negative charge, and the positive charge’s potential, therefore, will cancel with the negative charge’s potential because they’re going to be the same distance away to every point on this plane.

Another thing that we can conclude by looking at this example, since the positive charge’s potential is *q* over 4*π* *ε*0 *r* and the negative charge’s is *q* over –*q* over 4*π* *ε*0 *r*, therefore, if these two charges are same distance away to the point of interest, since positive is greater than negative, the potential associated with the positive charge is greater than the potential associated with the negative charge. But when we add these two, they will cancel one another on the equatorial plane of this dipole.

By the same token, we can see that when *θ* is equal to 0 degrees, and that is point along the dipole axis basically here to the positive charge, then the potential becomes *P* over 4*π* *ε*0 *r*2, since cosine of 0 is just 1. For *θ* is equal to 180 degrees, now we are near to the negative charge along the axis, then the potential is going to be –*P* over 4*π* *ε*0 *r*2 since cosine of 180 degrees is equal to -1. If we go back and look at our diagram then, when the angle *θ* is 0, we’re talking about along the axis of the dipole, the 0 will correspond to the points near to the positive charge, we end up with the net positive potential. For *θ* is equal to 180 degrees, now we are talking about near to the negative charge side along the axis, and then the net potential in that case is going to be negative, because the negative charge will be nearer to the point of interest relative to the positive charge.